Exercise 34.5

Solution. If $Y$ is metrizable, then it is second countable by Exercise 30.4. $X$ is then second countable by Theorem $30.2$.

Now suppose $X$ has a countable basis $B=\{B_i\}$. We claim that $B$ with sets of the form $Y\setminus \bigcup <!--\; FUNCTION\; APPLICATION\; -->\overline{B_i}$, where the union is finite and the closure is taken in $Y$, form a basis of $Y$; call this larger basis $B^{\text{+}}$. The $B_i\in B$ are open by Lemma $16.2$, and the $Y\setminus \bigcup <!--\; FUNCTION\; APPLICATION\; -->\overline{B_i}$ by Theorem $26.3$. $B^{\text{+}}$ is countable since $B$ is countable and there are countably many $Y\setminus \bigcup <!--\; FUNCTION\; APPLICATION\; -->\overline{B_i}$ by Exercise $\text{(}j)$.

Now recall that we have two types of open sets of $Y$, sets $U$ open in $X$, and sets $Y\setminus C$ for $C$ compact in $X$, by construction in Theorem $29.1$. By Lemma $13.2$, it suffices to show that for each $x$ in an open set, we can find an element of $B^{\text{+}}$ containing $x$ properly contained in the open set. So first consider the $U$. For every $x\in U$, there exists $B_i\in B$ such that $x\in B_i\subseteq U$ since $B$ is a basis of $X$. Now consider the $Y\setminus C$ and $x\in Y\setminus C$. If $x\in X$, then $Y\setminus C\cap X\ni x$ is open in $X$ by Theorem $29.1$ (which says $X$ is a subspace), and so the previous argument for $U\subseteq X$ applies. It remains to show the case $x=\infty$, where $\{\infty \}=Y\setminus X$. For each $y\in C$, we can find a neighborhood $V\ni y$ in $X$ such that $\overline{V}\subseteq X=Y\setminus \{x\}$ by local compactness. Since $V$ is open in $X$, there exists some $B_i\subseteq V$ containing $y$. These $B_i$ cover $C$ and so there is a finite subcover by compactness $C\subseteq \bigcup <!--\; FUNCTION\; APPLICATION\; -->B_i$; also, $x\notin \bigcup <!--\; FUNCTION\; APPLICATION\; -->\overline{B_i}\supseteq C$. Thus, we have $x\in Y\setminus \bigcup <!--\; FUNCTION\; APPLICATION\; -->\overline{B_i}\subseteq Y\setminus C$. $B^{\text{+}}$ is therefore a basis for $Y$ by Lemma $13.2$.

Since $Y$ is compact and Hausdorff, it is normal by Theorem 32.3, and in particular, $Y$ is regular. Finally, since $Y$ also has a countable basis, it is metrizable by the Urysohn metrization theorem (Theorem $34.1$). □